Transcript
CE 341/441 - Lecture 6 - Fall 2004
LECTURE
LAGRANG
Fit
= ex d to estab-lish an in
= a hrough allspecified ).
N 1+
f x( )
g x( )p. 6.1
6
E INTERPOLATION
points with an degree polynomial
act function of which only discrete values are known and useterpolating or approximating function
pproximating or interpolating function. This function will pass t interpolation points (also referred to as data points or nodes
Nth
f1
x0
g(x)f(x)
f0
f2f3 f4 fN
x1 x2 x3 x4 xN...
N 1+g x( )
N 1+
CE 341/441 - Lecture 6 - Fall 2004
The inter
There exi et ofpoints. It
where
No matter
Fitting
Lagra
Newto
The resul
N 1+
aip. 6.2
polation points or nodes are given as:
:
sts only one degree polynomial that passes through a given ss form is (expressed as a power series):
= unknown coefficients, ( coefficients).
how we derive the degree polynomial,
power series
nge interpolating functions
n forward or backward interpolation
ting polynomial will always be the same!
xo f xo( ) f ox1 f x1( ) f 1x2 f x2( ) f 2
xN f xN( ) f N
Nth
g x( ) ao a1x a2x2 a3x3 aN xN+ + + + +=
i 0 N,= N 1+
Nth
CE 341/441 - Lecture 6 - Fall 2004
Power Seri
mus
Solve set
It is relati ating func-tion
g x( )
g x( )p. 6.3
es Fitting to Define Lagrange Interpolation
t match at the selected data points
: :
of simultaneous equations
vely computationally costly to solve the coefficients of the interpol (i.e. you need to program a solution to these equations).
f x( )
g xo( ) f o= ao a1xo a2xo2 aN xoN+ + + + f o=
g x1( ) f 1= ao a1x1 a2x12+ + aN x1N+ + f 1=
g xN( ) f N= ao a1xN+ a2xN2 aN xNN+ + + f N=
1 xo xo2
xoN
1 x1 x12
x1N
1 xN xN2
xNN
ao
a1
:
aN
f of 1:
f N
=
CE 341/441 - Lecture 6 - Fall 2004
Lagrange I
We note t
Let
where at
For exam
Using the ;
Vi
3 x3( ) 1=V 4 x3( ) =p. 6.4
nterpolation Using Basis Functions
hat in general
= polynomial of degree associated with each node such th
ple if we have 5 interpolation points (or nodes)
definition for : ; ; ;,we have:
g xi( ) f i=
g x( ) f i Vi x( )i 0=
N
=
x( ) N i
Vi x j( )0 i j1 i = j
g x3( ) f oV o x3( ) f 1V 1 x3( ) f 2V 2 x3( ) f 3V 3 x3( ) f 4V 4 x3( )+ + + +=
V i x j( ) V 0 x3( ) 0= V 1 x3( ) 0= V 2 x3( ) 0= V0
g x3( ) f 3=
CE 341/441 - Lecture 6 - Fall 2004
How do w
Degre
Roots
Let
The f als zero atnode
Degre
Howe
We norma
V i xi( )
W i x( )p. 6.5
e construct ?
e
at (at all nodes except )
unction is such that we do have the required roots, i.e. it equs except at node
e of is
ver in the form presented will not equal to unity at
lize and define the Lagrange basis functions
V i x( )
N
xo x1 x2 xi 1 xi 1+ xN, , , , , , xi
1=
x xo( ) x x1( ) x x2( ) x xi 1( ) x xi 1+( ) x xN( )=
W ixo x1 x2 ... , xN, , , xi
W i x( ) N
W i x( ) xi
W i x( ) V i x( )
Vi x( )x xo( ) x x1( ) x x2( ) x xi 1( ) x xi 1+( ) x xN( )
xi xo( ) xi x1( ) xi x2( ) xi xi 1( ) xi xi 1+( ) xi xN( )-----------------------------------------------------------------------------------------------------------------------------------------------------------=
CE 341/441 - Lecture 6 - Fall 2004
Now we h
We also s
The gener f is:
The su
.
V i
V i x( )
g x( ) o aN, ,p. 6.6
ave such that equals:
atisfy
e.g.
al form of the interpolating function with the specified form o
m of polynomials of degree is also polynomial of degree
is equivalent to fitting the power series and computing coefficients
V i x( ) V i xi( )
xi( )xi xo( ) xi x1( ) xi x2( ) xi xi 1( ) 1( ) xi xi 1+( ) xi xN( )
xi xo( ) xi x1( ) xi x2( ) xi xi 1( ) xi xi 1+( ) xi xN( )--------------------------------------------------------------------------------------------------------------------------------------------------------------=
V i xi( ) 1=
V i x j( ) 0 for i j=
V 1 x2( )x2 xo( ) 1( ) x2 x2( ) x2 x3( ) x2 xN( )x1 xo( ) 1( ) x1 x2( ) x1 x3( ) x1 xN( )
---------------------------------------------------------------------------------------------------------- 0= =
g x( )
g x( ) f iVi x( )i 0=
N
=
N N
a
CE 341/441 - Lecture 6 - Fall 2004
Lagrange L
Linear La
wherep. 6.7
inear Interpolation Using Basis Functions
grange is the simplest form of Lagrange Interpolation
and
N 1=( )
g x( ) f iV i x( )i 0=
1
=
g x( ) f oV o x( ) f 1V 1 x( )+=
V o x( )x x1( )xo x1( )
---------------------
x1 x( )x1 xo( )
---------------------= = V 1 x( )x xo( )x1 xo( )
---------------------=
x0(x)
V0 (x)
x1
V1(x)1.0
CE 341/441 - Lecture 6 - Fall 2004
Example
Given the
Find the l
Lagrange
Interpolatp. 6.8
following data:
inear interpolating function
basis functions are:
and
ing function g(x) is:
xo 2= f o 1.5=x1 5= f 1 4.0=
g x( )
V o x( )5 x
3-----------= V 1 x( )x 2
3-----------=
g x( ) 1.5V o x( ) 4.0V 1 x( )+=
CE 341/441 - Lecture 6 - Fall 2004p. 6.9
x0 = 2
1.5 V0 (x)42
42
x1 = 5x
4.0 V1(x)
x0 = 2 x1 = 5x
x0 = 2 x1 = 5
g(x) = 1.5 V0(x) + 4.0V1(x)
CE 341/441 - Lecture 6 - Fall 2004
Lagrange Q For quadr
wherep. 6.10
uadratic Interpolation Using Basis Functionsatic Lagrange interpolation, N=2
g x( ) f i V i x( )i 0=
2
=
g x( ) f oV o x( ) f 1V 1 x( ) f 2V 2 x( )+ +=
V o x( )x x1( ) x x2( )
xo x1( ) xo x2( )-------------------------------------------=
V 1 x( )x xo( ) x x2( )
x1 xo( ) x1 x2( )-------------------------------------------=
V 2 x( )x xo( ) x x1( )
x2 xo( ) x2 x1( )-------------------------------------------=
CE 341/441 - Lecture 6 - Fall 2004
Note that ch that thebasic prep. 6.11
the location of the roots of , and are defined sumise of interpolation is satisfied, namely that . Thus:
x0
V0 (x)1.0
x1 x
x2
V1(x) V2(x)
V 0 x( ) V 1 x( ) V 2 x( )g xi( ) f i=
g xo( ) V o xo( ) f o V 1 xo( ) f 1 V 2 xo( ) f 2+ + f 0= =
g x1( ) V o x1( ) f o V 1 x1( ) f 1 V 2 x1( ) f 2+ + f 1= =
g x2( ) V o x2( ) f o V 1 x2( ) f 1 V 2 x2( ) f 2+ + f 2= =
CE 341/441 - Lecture 6 - Fall 2004
Example Given the
Find the q
Lagrange
Interpolatp. 6.12
following data:
uadratic interpolating function
basis functions are
ing function g(x) is:
xo 3= f o 1=x1 4= f 1 2=x2 5= f 2 4=
g x( )
V o x( ) x 4( ) x 5( )3 4( ) 3 5( )----------------------------------=
V 1 x( ) x 3( ) x 5( )4 3( ) 4 5( )----------------------------------=
V 2 x( ) x 3( ) x 4( )5 3( ) 5 4( )----------------------------------=
g x( ) 1.0V o x( ) 2.0V 1 x( ) 4.0V 2 x( )+ +=
CE 341/441 - Lecture 6 - Fall 2004p. 6.13
1.0 V0 (x)1.0x
4.0 V2(x)
x1 = 4x0 = 3 x2 = 5
2.0
4.0
x1 = 4x0 = 3 x2 = 5
x1 = 4x0 = 3 x2 = 5
x
x
2.0 V1(x)
x1 = 4x0 = 3 x2 = 5
4.0g(x) = 1.0 V0(x) + 2.0V1(x) + 4.0V2(x)
CE 341/441 - Lecture 6 - Fall 2004
Lagrange C
For Cubic
Example Consider )
Find g 0.(p. 6.14
ubic Interpolation Using Basis Functions
Lagrange interpolation, N=3
the following table of functional values (generated with
as:
0 0.40 -0.9162911 0.50 -0.6931472 0.70 -0.3566753 0.80 -0.223144
f x( ) xln=
i xi f i
60)
g x( ) f ox x1( ) x x2( ) x x3( )
xo x1( ) xo x2( ) xo x3( )----------------------------------------------------------------- f 1
x xo( ) x x2( ) x x3( )x1 xo( ) x1 x2( ) x1 x3( )
-----------------------------------------------------------------+=
f 2x xo( ) x x1( ) x x3( )
x2 xo( ) x2 x1( ) x2 x3( )----------------------------------------------------------------- f 3
x xo( ) x x1( ) x x2( )x3 xo( ) x3 x1( ) x3 x2( )
-----------------------------------------------------------------+ +
CE 341/441 - Lecture 6 - Fall 2004p. 6.15
g 0.60( ) 0.916291 0.60 0.50( ) 0.60 0.70( ) 0.60 0.80( )0.40 0.50( ) 0.40 0.70( ) 0.40 0.80( )------------------------------------------------------------------------------------------------=
0.693147 0.60 0.40( ) 0.60 0.70( ) 0.60 0.80( )0.50 0.40( ) 0.50 0.70( ) 0.50 0.80( )------------------------------------------------------------------------------------------------
0.356675 0.60 0.40( ) 0.60 0.50( ) 0.60 0.80( )0.70 0.40( ) 0.70 0.50( ) 0.70 0.80( )------------------------------------------------------------------------------------------------
0.223144 0.60 0.40( ) 0.60 0.50( ) 0.60 0.70( )0.80 0.40( ) 0.80 0.50( ) 0.80 0.70( )------------------------------------------------------------------------------------------------
g 0.60( ) 0.509976=
CE 341/441 - Lecture 6 - Fall 2004
Errors Ass
Using Tay
where
Notes
If
Theref
f N +
L x(
f x(p. 6.16
ociated with Lagrange Interpolation
lor series analysis, the error can be shown to be given by:
derivative of w.r.t. evaluated at
= polynomial of degree where , then
for all xore will be an exact representation of
e x( ) f x( ) g x( )=
e x( ) L x( ) f N 1+( ) ( )= xo xN
1 ( ) N 1th+= f x
) x xo( ) x x1( ) x xN( )N 1+( )!----------------------------------------------------------------- an N 1
th degree polynomial+= =
) M M N
f N 1+( ) x( ) 0= e x( ) 0=g x( ) f x( )
CE 341/441 - Lecture 6 - Fall 2004
Sincedoes
es
e.g
e interval.
f N 1+( ) x( )
f N 1+(
L x( )
L x( )
L
Lp. 6.17
in general is not known, if the interval is small and ifnot change rapidly in the interval
where .
can be estimated by using Finite Difference (F.D.) formulae
will significantly effect the distribution of the error
is a minimum at the center of and a maximum near the edg
. using 6 point interpolation looks like:
at all data points
largest . becomes very large outside of th
xo xN,[ ]
e x( ) L x( ) f N 1+( ) xm( ) xmxo xN+
2-------------------=
)
xo xN,[ ]
L x( )
0 1 2 3 4 5
x( ) 0=x( ) 0 x 1 4 x 5 L x( )
CE 341/441 - Lecture 6 - Fall 2004
As the rror withinthe in
As
Howe
Th
Prope
N
xx0 x xN p. 6.18
size of the interpolating domain increases, so does the maximum eterval
increases from a small value,
ver as for a given and thus
erefore convergence as does not necessarily occur!!
rties of will also influence error as and vary
D xN xo= Lmaxx0 x xN
emaxx0 x xN
Lmaxx0 x xN
emaxx0 x xN
N NCRIT> Lmaxx0 x xN
xo xN,[ ] ema
N
f N 1+( ) ( ) D N
CE 341/441 - Lecture 6 - Fall 2004
Example
Estimate usually wedo not ha
e
)
ep. 6.19
the error made in the previous example knowing that (ve this information).
f x( ) x( )ln=
x( ) L x( ) f N 1+( ) xm( )
e x( ) x xo( ) x x1( ) x x2( ) x x3( )3 1+( )!----------------------------------------------------------------------------- f3 1+( )
xm( )
e 0.60( ) 0.60 0.40( ) 0.60 0.50( ) 0.60 0.70( ) 0.60 0.80( )3 1+( )!-------------------------------------------------------------------------------------------------------------------------------- f3 1+( ) 0.6(
0.60( ) 0.000017 f 4( ) 0.6( )=
CE 341/441 - Lecture 6 - Fall 2004
We estim
Therefore
Exact erro
Therefore
Typically nce (F.D.)approximp. 6.20
ate the fourth derivative of f(x) using the analytical function itself
r is computed as:
error estimate is excellent
we would also have to estimate using a Finite Differeation (a discrete differentiation formula).
f x( ) xln=
f 1( ) x( ) x 1=
f 2( ) x( ) x 2=
f 3( ) x( ) 2x 3=
f 4( ) x( ) 6x 4=
f 4( ) 0.6( ) 46.29=
e 0.60( ) 0.00079=
E x( ) 0.60( ) g 0.60( )ln 0.00085= =
f N 1+( ) xm( )
CE 341/441 - Lecture 6 - Fall 2004
SUMMAR
Linear int
Power se ficients bysolving a
Lagrange points Use spec
where
fi = th
Each L nity at thedata onzero in-betw
g x( )
V i x( )p. 6.21
Y OF LECTURES 5 AND 6
erpolation passes a straight line through 2 data points.
ries data points degree polynomial find coef matrix
Interpolation passes an degree polynomial through dataialized nodal functions to make finding easier.
= the interpolating function approximating f(x)e value of the function at the data (or interpolation) point i = the Lagrange basis function
agrange polynomial or basis function is set up such that it equals upoint with which it is associated, zero at all other data points and neen.
N 1+ Nth
Nth N 1+g x( )
g x( ) f iV i x( )i 0=
N
=
CE 341/441 - Lecture 6 - Fall 2004
For exp. 6.22
ample when N 2= 3 data points
V0 V1 V2
0 1 2
g x( ) f oV o x( ) f 1V 1 x( ) f 2V 2 x( )+ +=
f0
f1
f2 g(x)
CE 341/441 - Lecture 6 - Fall 2004
Linear int
Error esti e point inthe interv
where
f N 1+
L x( )p. 6.23
erpolation is the same as Lagrange Interpolation with
mates can be derived but depend on knowing (or at somal).
derivative of w.r.t. evaluated at
N 1=
f N 1+( ) xm( )
e x( ) L x( ) f N 1+( ) ( )= xo xN
( ) N 1th+= f x x xo( ) x x1( ) x xN( )
N 1+( )!----------------------------------------------------------------- an N 1th degree polynomial+= =
LECTURE 6LAGRANGE INTERPOLATION Fit points with an degree polynomial = exact function of which only discrete values are known and used to establish an interpolating... = approximating or interpolating function. This function will pass through all specified interp... The interpolation points or nodes are given as::
There exists only one degree polynomial that passes through a given set of points. Its form is...where = unknown coefficients, ( coefficients). No matter how we derive the degree polynomial, Fitting power series Lagrange interpolating functions Newton forward or backward interpolation
The resulting polynomial will always be the same!
Power Series Fitting to Define Lagrange Interpolation must match at the selected data points : :
Solve set of simultaneous equations It is relatively computationally costly to solve the coefficients of the interpolating function...
Lagrange Interpolation Using Basis Functions We note that in general Letwhere = polynomial of degree associated with each node such that For example if we have 5 interpolation points (or nodes)
Using the definition for : ; ; ; ; ,we have: How do we construct ? Degree Roots at (at all nodes except )
Let The function is such that we do have the required roots, i.e. it equals zero at nodes except at... Degree of is However in the form presented will not equal to unity at
We normalize and define the Lagrange basis functions Now we have such that equals:
We also satisfye.g.
The general form of the interpolating function with the specified form of is: The sum of polynomials of degree is also polynomial of degree is equivalent to fitting the power series and computing coefficients .
Lagrange Linear Interpolation Using Basis Functions Linear Lagrange is the simplest form of Lagrange Interpolation
whereand
Example Given the following data:Find the linear interpolating function Lagrange basis functions are:and
Interpolating function g(x) is:
Lagrange Quadratic Interpolation Using Basis Functions For quadratic Lagrange interpolation, N=2
where Note that the location of the roots of , and are defined such that the basic premise of interpo...
Example Given the following data:Find the quadratic interpolating function Lagrange basis functions are Interpolating function g(x) is:
Lagrange Cubic Interpolation Using Basis Functions For Cubic Lagrange interpolation, N=3
Example Consider the following table of functional values (generated with )
00.40-0.91629110.50-0.69314720.70-0.35667530.80-0.223144 Find as:
Errors Associated with Lagrange Interpolation Using Taylor series analysis, the error can be shown to be given by:wherederivative of w.r.t. evaluated at Notes If = polynomial of degree where , then for all x
Therefore will be an exact representation of Since in general is not known, if the interval is small and if does not change rapidly in the i...where .
can be estimated by using Finite Difference (F.D.) formulae will significantly effect the distribution of the error is a minimum at the center of and a maximum near the edges e.g. using 6 point interpolation looks like: at all data points largest . becomes very large outside of the interval.
As the size of the interpolating domain increases, so does the maximum error within the interval
As increases from a small value, However as for a given and thus Therefore convergence as does not necessarily occur!!
Properties of will also influence error as and vary
Example Estimate the error made in the previous example knowing that (usually we do not have this infor...
We estimate the fourth derivative of f(x) using the analytical function itself
Therefore Exact error is computed as:Therefore error estimate is excellent Typically we would also have to estimate using a Finite Difference (F.D.) approximation (a disc...
SUMMARY OF LECTURES 5 AND 6 Linear interpolation passes a straight line through 2 data points. Power series data points degree polynomial find coefficients by solving a matrix Lagrange Interpolation passes an degree polynomial through data points Use specialized nodal ...where= the interpolating function approximating f(x)fi = the value of the function at the data (or interpolation) point i= the Lagrange basis function Each Lagrange polynomial or basis function is set up such that it equals unity at the data poin... For example when Linear interpolation is the same as Lagrange Interpolation with Error estimates can be derived but depend on knowing (or at some point in the interval).
wherederivative of w.r.t. evaluated at
top related